$$\begin{aligned} i &= 0 + i = \cos(\frac{\pi}{2}) + i \sin(\frac{\pi}{2})\\ &= e^{\frac{i \pi}{2}} \end{aligned}$$

Thanks to Euler! Therefore,

$$\begin{aligned} i^i &= (e^{\frac{i \pi}{2}})^i = e^{\frac{i^2 \pi}{2}} \\ &= e^{\frac{-\pi}{2}} = \frac{1}{e^{\frac{\pi}{2}}} \\ &\approx 0.2078796 \end{aligned}$$

Furthermore, there is apparently more than one solution, as $i^i$ can be derived with multiples of $2 \pi$ in the periodic functions. The solution set can therefore be represented by a sequence, with

$$\begin{aligned} a_n = e^{-(\frac{\pi}{2} + 2\,n\,\pi)} \end{aligned}$$

where $n$ is any integer.

For example, for $n = -1,0,1,\cdots$, etc. $a_n = 111.3178, 0.2078796, 0.0003882032, \cdots$, etc.

# posted by rot13(Unafba Pune) @ 5:08 PM